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  1. Prepare
  2. Algorithms
  3. Implementation
  4. 3D Surface Area
  5. Discussions

3D Surface Area

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  • naufalfh19
     + 0 comments

    Python3 Solution

    def surfaceArea(A):
    res = 0
    row = len(A)
    col = len(A[0])
    
    # calculate front and back
    for i in range(row):
        for j in range(col):
            if j != 0:
                res += max(0, A[i][j]-A[i][j-1])
            else:
                res += A[i][j]
            
            if j != col-1:
                res += max(0, A[i][j]-A[i][j+1])
            else:
                res += A[i][j]
                
            if i != 0:
                res += max(0, A[i][j]-A[i-1][j])
            else:
                res += A[i][j]
                
            if i != row-1:
                res += max(0, A[i][j]-A[i+1][j])
            else:
                res += A[i][j]
 
    res += 2*(row*col)       
    
    return res

  • wifivop476
     + 0 comments

    I approached this problem counting surface from 4 directions: front, rear, left and right. Of course top and bottom are just HW. While thinking about it, perhaps you noticed that rather than 4 directions, we only need 2 by using the "abs" function, similar to how services for BMW Service Near Me optimize calculations to efficiently manage deliveries. Brilliant observation mate! So well done!

  • wifivop476
     + 0 comments

    I approached this problem counting surface from 4 directions: front, rear, left and right. Of course top and bottom are just HW. While thinking about it, perhaps you noticed that rather than 4 directions, we only need 2 by using the "abs" function, similar to how services like adas calibration near me optimize calculations to efficiently manage deliveries. Brilliant observation mate! So well done!

  • yashdeora98294
     + 0 comments

    Here is problem solution in python, java, c++, c and javascript programming - https://programmingoneonone.com/hackerrank-3d-surface-area-problem-solution.html

  • ardhian777
     + 0 comments

    simulate putting each cube, for each cube, check 6 sides, if a cube exist on a side minus 2

    def surfaceArea(A):
    H = len(A)
    W = len(A[0])
    cubemap = set()
    tot = 0

    # Directions for 6 neighbors: (di, dj, dk)
    directions = [(1,0,0), (-1,0,0), (0,1,0), (0,-1,0), (0,0,1), (0,0,-1)]

    for i in range(H):
        for j in range(W):
            for k in range(A[i][j]):
                tot += 6  # Each cube starts with 6 faces
                # Check 6 neighbors
                for di, dj, dk in directions:
                    ni, nj, nk = i + di, j + dj, k + dk
                    if (ni, nj, nk) in cubemap:
                        tot -= 2  # Shared face
                cubemap.add((i, j, k))

    return tot